Question

1. Consider the following reaction: 2 Al + 6 HBr → 2 AlBr3 + 3 H2 

1. What is the limiting reactant when 3.22 moles of Al reacts with 4.96 moles of HBr? 
2. How many moles of H2 are formed?

Answer 1

$1)\ 2Al \ +$$6HBr\to$ $2AlBr_3+3H_2$

Since 2 moles aluminium are required for every 6 moles of hydrogen bromide, yet we only have 3.22 moles aluminium and 4.96 moles of hydrogen bromide , it implies that HB$r$ will get used up first and hence it is limiting reactant while Al is in excess.

So all 4.96 mol of HBr is used up and reacts with$\ =\dfrac{4.96}{3}=1.653 \ mol Al$ to produce$=\dfrac{4.96}{2}=2.48\ mol H_2$